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Especially Lime
Especially Lime
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Suppose there are $R$ red lamps and $150-R$ blue.

For a given blue lamp, suppose there are $x$ red lamps earlier in the ordering (so $R-x$ later). Then that blue lamp is in $x(R-x)$ scoring triples.

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Now $$x(R-x)=(R/2-(R/2-x))(R/2+(R/2-x))=R^2/4-(R/2-x)^2,$$ so this is maximised (for a fixed value of $R$) when $x$ is as close as possible to $R/2$. We can achieve this maximum for every blue lamp simultaneously by having $\lfloor R/2\rfloor$ red lamps, then $150-R$ blue lamps, then $\lceil R/2\rceil$ red lamps.

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This gives a score of $(150-R)\lfloor R^2/4\rfloor\leq (150-R)R^2$. Using AM-GM with $150-R,R/2,R/2$, this is maximised when $R/2=150-R$, i.e. $R=100$. In this case since $R$ is even we attain the upper bound, so 50 red, 50 blue, 50 red is the best option.

Suppose there are $R$ red lamps and $150-R$ blue.

For a given blue lamp, suppose there are $x$ red lamps earlier in the ordering (so $R-x$ later). Then that blue lamp is in $x(R-x)$ scoring triples.

.

Now $$x(R-x)=(R/2-(R/2-x))(R/2+(R/2-x))=R^2/4-(R/2-x)^2,$$ so this is maximised (for a fixed value of $R$) when $x$ is as close as possible to $R/2$. We can achieve this maximum for every blue lamp simultaneously by having $\lfloor R/2\rfloor$ red lamps, then $150-R$ blue lamps, then $\lceil R/2\rceil$ red lamps.

.

This gives a score of $(150-R)\lfloor R^2/4\rfloor\leq (150-R)R^2$. Using AM-GM with $150-R,R/2,R/2$, this is maximised when $R/2=150-R$, i.e. $R=100$. In this case since $R$ is even we attain the upper bound, so 50 red, 50 blue, 50 red is the best option.

Suppose there are $R$ red lamps and $150-R$ blue.

For a given blue lamp, suppose there are $x$ red lamps earlier in the ordering (so $R-x$ later). Then that blue lamp is in $x(R-x)$ scoring triples. 

Now $$x(R-x)=(R/2-(R/2-x))(R/2+(R/2-x))=R^2/4-(R/2-x)^2,$$ so this is maximised (for a fixed value of $R$) when $x$ is as close as possible to $R/2$. We can achieve this maximum for every blue lamp simultaneously by having $\lfloor R/2\rfloor$ red lamps, then $150-R$ blue lamps, then $\lceil R/2\rceil$ red lamps. 

This gives a score of $(150-R)\lfloor R^2/4\rfloor\leq (150-R)R^2$. Using AM-GM with $150-R,R/2,R/2$, this is maximised when $R/2=150-R$, i.e. $R=100$. In this case since $R$ is even we attain the upper bound, so 50 red, 50 blue, 50 red is the best option.

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Especially Lime
Especially Lime
  • 3.2k
  • 12
  • 17

Suppose there are $R$ red lamps and $150-R$ blue.

For a given blue lamp, suppose there are $x$ red lamps earlier in the ordering (so $R-x$ later). Then that blue lamp is in $x(R-x)$ scoring triples.

.

Now $$x(R-x)=(R/2-(R/2-x))(R/2+(R/2-x))=R^2/4-(R/2-x)^2,$$ so this is maximised (for a fixed value of $R$) when $x$ is as close as possible to $R/2$. We can achieve this maximum for every blue lamp simultaneously by having $\lfloor R/2\rfloor$ red lamps, then $150-R$ blue lamps, then $\lceil R/2\rceil$ red lamps.

.

This gives a score of $(150-R)\lfloor R^2/4\rfloor\leq (150-R)R^2$. Using AM-GM with $150-R,R/2,R/2$, this is maximised when $R/2=150-R$, i.e. $R=100$. In this case since $R$ is even we attain the upper bound, so 50 red, 50 blue, 50 red is the best option.